numpy.random.exponential

random.exponential(scale=1.0, size=None)

Draw samples from an exponential distribution.

Its probability density function is

\[f(x; \frac{1}{\beta}) = \frac{1}{\beta} \exp(-\frac{x}{\beta}),\]

for x > 0 and 0 elsewhere. \(\beta\) is the scale parameter, which is the inverse of the rate parameter \(\lambda = 1/\beta\). The rate parameter is an alternative, widely used parameterization of the exponential distribution [3].

The exponential distribution is a continuous analogue of the geometric distribution. It describes many common situations, such as the size of raindrops measured over many rainstorms [1], or the time between page requests to Wikipedia [2].

Note

New code should use the exponential method of a Generator instance instead; please see the Quick start.

Parameters:

scalefloat or array_like of floats

The scale parameter, \(\beta = 1/\lambda\). Must be non-negative.

sizeint or tuple of ints, optional

Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if scale is a scalar. Otherwise, np.array(scale).size samples are drawn.

Returns:

outndarray or scalar

Drawn samples from the parameterized exponential distribution.

See also

random.Generator.exponential

which should be used for new code.

References

[1]

Peyton Z. Peebles Jr., “Probability, Random Variables and Random Signal Principles”, 4th ed, 2001, p. 57.

[2]

Wikipedia, “Poisson process”, https://en.wikipedia.org/wiki/Poisson_process

[3]

Wikipedia, “Exponential distribution”, https://en.wikipedia.org/wiki/Exponential_distribution

Examples

Try it in your browser!

A real world example: Assume a company has 10000 customer support agents and the average time between customer calls is 4 minutes.

>>> n = 10000
>>> time_between_calls = np.random.default_rng().exponential(scale=4, size=n)

What is the probability that a customer will call in the next 4 to 5 minutes?

>>> x = ((time_between_calls < 5).sum())/n
>>> y = ((time_between_calls < 4).sum())/n
>>> x-y
0.08 # may vary

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